double comb space is not contractible

I'm trying to show that the double comb space is not contractible.

Intuitively I can see why this is true, but I can't seem to formalize a prof.

I try to do the following:

Let $D$ be the double comb space Suppose $H:D\times I \rightarrow D$ so that $H(x,0)=x$ and $H(x,1)=x_0$ where $x_0\in D$

$D$ is path-connected so we can assume $x_0=(0,0)$.

Now we need to somehow show that H is not continuous, I'm just not sure how.

I have a feeling that I'm not understanding some basic idea here, and that this prof should be quite simple, but I'm not sure what it might be.

Thanks.

• this looks very similar to this question math.stackexchange.com/questions/22002/… Jun 18 '13 at 15:12
• I am not sure that we can assume $x_0=(0,0)$ just because the space is path connected. E.g. even the single comb space is path connected and deformation retracts to any point in the base, but not to the end of the last bristle. Jan 18 '15 at 15:00
• I'm not assuming that the space deformation retracts to $x_0$, but that the constant function equal to $x_0$ is homotopic to the identity function. Since in a path connected space any two constant functions are homotopic, I can assume $x_0=(0,0)$. Jan 19 '15 at 15:32

Have you tried using uniform continuity? $D \times I$ is a compact metric space so the function $H$, if it is continuous, must also be uniformly continuous. This could have implications for how $H$ maps two sets $x \times I$ and $y \times I$ when $x,y \in D$ are very close to $x_0$ but $x$ is to the left and $y$ is to the right of $x_0$.

Let $x_0=(0,1)$, the point where the two combs are wedged.

There is $t_0$ such that

$H(x_0\times[0,t_0])=\{x_0\}$ and for each $\epsilon>0$ there is an $0<\alpha<\epsilon$ and a $\lambda\ne0$ with $H(x_0,\ t_0+\alpha)=(0,\ 1+\lambda)$

Take $0<\epsilon<1$. By continuity, there is a $\delta>0$ with $$H(B_δ(x_0)×[0,\ t_0+δ])\subseteq B_ϵ(x_0)$$ There is an $0<\alpha<δ$ with $$H(x_0,\ t_0+α)=(0,\ 1+\lambda)$$ where $λ\in(-ϵ,ϵ)$. Again by continuity, there is a $0<\beta<α$ such that $$H(B_β(x_0)×[-β+(t_0+α),\ (t_0+α)+β])\subseteq B_λ(0,\ 1+λ)$$ Assume without loss of generality that $λ>0$. But what does this imply for a point $y=(y_1,1)$ where $0<y_1<β$. It means that $y$ must travel along a path to a point in the upper comb (since $B_λ(0,1+λ)$ doesn't meet the lower comb), where it arrives at some time $t\in((t_0+α)-β,\ (t_0+α)+β)$. To do so, it had to go down to $I×\{0\}$, which is impossible since $H(B_δ(x_0)×[0,\ t_0+δ])⊆B_ϵ(x_0)$.

There is still the possibility that $t_0=1$ and the point $(0,1)$ is fixed during the $I$. But that would mean that the comb space strongly deformation retracted onto this point. But then for some small ball $B$ around $x_0$, some smaller ball had to be stay in $B$ during the entire time, preventing the point $y_1$ from travelling along the path through the base.

• Sorry I do not understand why there should be a $t_0$ satisfying the criteria. In other words, why is it that $x_0$ can not start moving instantly ? Jan 18 '15 at 15:03
• @user90041: The $t_0$ is computed as follows: We have a map $t\mapsto H(x_0,t)$, which sends $0$ to $x_0$. Since $\{x_0\}$ is closed, $H(x_0,-)^{-1}(\{x_0\})$ is a closed compact subset of $I$ containing $0$. The connected component of this set which contains $0$ is the interval $[0,t_0]$. Of course, $t_0$ could just be $0$. Jan 19 '15 at 2:26